Clearly, the existence of \({\mathcal {M}}(f, \mu , \nu )\)  is guaranteed by Theorem REF . Without lost of generality, suppose \(a > 0\) . Now since \(f \in AP(\mathbb {X})\)  it follows that \(f_a: t \mapsto f(t + a)\)  belongs to \(AP(\mathbb {X})\) . Moreover, the weight \(\nu _a\)  defined by \(\nu _a(t) = \nu (t+a)\)  for all \(t \in \mathbb {R}\)  belongs to \(\mathbb {U}_\infty ^0\) .
Now
\(\Big | \int _{Q_T} e^{i \lambda t} \nu _a (t) dt \Big | = \Big | \int _{Q_T-a} e^{i \lambda (t-a)} \nu (t) dt \Big | = \Big | \int _{Q_T-a} e^{i \lambda t} \nu (t) dt \Big | \le \Big | \int _{Q_{T+a}} e^{i \lambda t} \nu (t) dt \Big |\) 
and hence
\(\lim _{T \rightarrow \infty } \Big | \frac{1}{\mu (Q_T)}\int _{Q_T} e^{i \lambda t} \nu _a (t) dt \Big | &=& \lim _{T\rightarrow \infty } \Big |\frac{1}{\mu (Q_{T})} \int _{Q_{T}-a} e^{i \lambda t} \nu (t) dt \Big | \\&\le & \lim _{T\rightarrow \infty } \Big |\frac{1}{\mu (Q_{T})} \int _{Q_{T+a}} e^{i \lambda t} \nu (t) dt \Big | \\&=& \lim _{T\rightarrow \infty } \Big |\frac{\mu (Q_{T+a})}{\mu (Q_{T})}\frac{1}{\mu (Q_{T+a})} \int _{Q_{T+a}} e^{i \lambda t} \nu (t) dt \Big | \\&=& D_a\lim _{T\rightarrow \infty } \Big |\frac{1}{\mu (Q_{T+a})} \int _{Q_{T+a}} e^{i \lambda t} \nu (t) dt \Big | \\&=&0.\) 
Now
\(\displaystyle \lim _{T \rightarrow \infty } \frac{\nu _a (Q_T)}{\mu (Q_T)} = C_{-a} \theta _{\mu \nu }.\) 
Using Theorem REF  it follows that for every \(\varphi \in AP(\mathbb {X})\) ,
\({\mathcal {M}}(\varphi _a, \mu , \nu _a) = \lim _{T \rightarrow \infty } \frac{1}{\mu (Q_T)} \int _{Q_T} \varphi _a(t) \nu _a(t) dt\) 
exists. Furthermore, \({\mathcal {M}}(\varphi _a, \mu , \nu _a) = C_{-a} \theta _{\mu \nu } {\mathcal {M}}(\varphi _a)\)  for all \(a \in \mathbb {R}\) . In particular, \({\mathcal {M}}(f_a, \mu , \nu _a) = C_{-a} \theta _{\mu \nu } {\mathcal {M}}(f_a)\)  uniformly in \(a \in \mathbb {R}\) . Now from Bohr [1]}, \({\mathcal {M}}(f_a) = {\mathcal {M}}(f)\)  uniformly in \(a \in \mathbb {R}\) , which completes the proof.
